00:00 - 19:0019:00 - 00:00

12:00 AM
I told you above. Looking at the picture is fine if you can see principal normals.

okay, I'll get to the Frenet normals soon.

12:43 AM
ok, my brain is fried right now

Is that a delicacy?

I have linear maps $f\colon A\rightarrow V$, $g\colon B\rightarrow V$ and $h\colon C\rightarrow V$ such that $(f,g)\colon A\oplus B\rightarrow V$ and $(g,h)\colon B\oplus C\rightarrow V$ are isomorphisms, does the image of $A\oplus B\oplus C\rightarrow V\oplus V,\,(a,b,c)\mapsto(f(a)+g(b),g(b)+h(c))$ and the diagonal together generate $V\oplus V$?
@TedShifrin I wish

So take $B$ to be the inclusion and $A$ and $C$ different complements. Do dimensions check?
Oh, generate.

yeah, the sum of these subspaces will actually never be direct for dimension reasons unless we are in the degenerate case where $A=C=0$

But the maps have to be injections, so isn’t this just about transversality of subspaces?

12:52 AM
yeah (this is actually a transversality computation for the trisecant map, which I thought I did a couple days ago, but now I'm not seeing it anymore)
ah, I get it now
my assumptions are not strong enough
I need to also assume that $(f,h)\colon A\oplus C\rightarrow V$ is an isomorphism
if I identify $V\oplus V/\Delta_V\cong V$ via $[v,w]\mapsto v-w$, all I need surjectivity of the map $A\oplus B\oplus C\rightarrow V,\,(a,b,c)\mapsto f(a)-h(c)$
so instead of just defining a trisecant map on the complement of $\Delta_M\times M$ and $M\times\Delta_M$ in $M\times M\times M$, I also have to exclude the points where the first and third coordinate are equal!

Yes, all three diagonals …

1:15 AM
I wonder, do people have a preferred notion for $X^n$ without all the diagonals

1:38 AM
on my phone but how can I use the pigeon hole principle to prove : For a given natural number a, there exists k and l such that a^k - a^l is divisible by 10. Im having trouble following the logic
How can we assume f:NxN to N, just because N^2 is "larger" than N or the same size then you have f a surjection?

@Obliv a^k has 10 possible last digits, so if you vary k 11 times then....
@Obliv What is f?

then? What if a is 2? a^k has 5 possible last digits then
f is the function that maps k,l to a^k-a^l, which turns out to be divisible by 10? apparently

I have a very silly way to verify this
just write the powers from 1 to 9 and check how they regress
for example any number with the last digit 2, and you take the power of it you will get the last digit 2, 4, 8, 6, 2

@Obliv I mean at most 10 possible last digits, so if you input 11 values of k then at least two of the values will have same last digit

yep, pigeon-hole works

1:55 AM
Well actually at most 5 possible last digits

yeah it depends if a is even or odd
ud only need 6 different k to have a repeat for either a even or odd

can u explain like im 5 years old how that proves a^k - a^l is divisible by 10 for some k,l in N

I guess what he means is that whatever a is and for whatever the last digit a has, by taking multiple power you will eventually obtain a number that has the same last digit and in which case their differences mod 10.

ohhh like a^k-a^l mod 10

2:00 AM
yes
like 11, $11^2=121$ and $121-11\equiv 0 \pmod 10$ something like this

i still dont get why we use the pigeon hole principle. Why not just prove it directly
like if a even or a odd, we can literally just have that difference equiv 0 mod 10
How would the pigeon hole principle way go?

using pigeon principle you then not need to worry about the specific last digit anymore, all you care is that there are only 10 possible last digits whenever you are taking the power. And in the 11th time, you will have at least two powers have the same last digit which implies ....

@Obliv Then you have to write down enough possible powers of $0,1,...,9$ to show it is true

can we use induction?

induction like?

2:05 AM
like base case is a = 1, then u say this works for any a, then a+1. Do cases of a even or odd, perhaps show that the last digits mod 10 must be in some set
and then show a^k and a^l uses the same set for last digits
a+1**
@oscarmetalbreak Is this then the extreme case where N^2 is larger than 10 so it must be a surjection
mapping NxN to {1,2,3,...,10} lol
or rather 0-9 for mod 10

But you have to introduce a formula, aka a way to prove they have the last digit, don't you in order to do induction?
but pigeon hole is sufficient don't you think?

yes it would be a^k-a^l=10b for some b in Z or u could do in mod 10 so =m such that m is 0-9
oh yeah if we do mod 10 we are mapping NxN to a set of size 10 lol
its very sufficient
I think i set it up wrong tho
Im pretty sure I should have done a^k-a^l = c for some c in Z

2:51 AM
no pigeons were hurt during the computation.

3:11 AM
Is there a relation between the existence of roots of the minimal polynomial to the matrix of a specific order that you can find? For example, Let $F$ be a field with prime number $4k+3$ of elements, then a matrix of order 4 in $GL_2(F)$ is unique up to conjugation, but the polynomial $x^4=1$ has no roots other than 1, -1. On the other hand when $F$ a field with prime number $4k+1$ of elements you have 7 such conjugate classes.
Maybe this is a simple linear algebra question but I really forgot...

sry, i only do complex

ironically, complex is simpler (for me at least)

Really? So you have an answer to this question if it was the complex field?
I don't mind to hear it though

it has no relevance that i can think of in the context of your question.
at the risk of inviting invective, you should ignore me...

3:28 AM
No worry. I guess maybe there is no general relation and the answer just depends from cases to cases.

3:39 AM
@oscarmetalbreak Think for starters about possible eigenvalues of such a matrix.
This is related to yesterday’s Wilson’s Theorem discussion.

3:56 AM
@SoumikMukherjee have you taken your interviews?
what did they ask you in TIFR interview which you taken in past. I mean it was full of mathematics or other things as well?

4:11 AM
@TedShifrin I think I got it but can you confirm my arguement? Since we are looking for a matrix $A$ satisfy $x^4-1=0$, which means the minimal polynomial of $A$ must divide that. In the first case, $p=4k+3$, one can write $x^4-1=(x^2+1)(x^2-1)$, and (x^2+1) is the only option of the minimal polynomial since otherwise A will be order of 2. However, when $p=4k+1$, one can write $x^4-1=(x+ p)(x-q)(x-1)(x+1)$ or $(x- p)(x+q)(x-1)(x+1)$ because of Wilson's theorem.
In the latter case, we can find the possible minimal polynomial of degree 2 by combining the linear factors and excluding the repeated ones and $x^2-1$, we will get 7.
plus $x^2+1$ for sure
I feel something wrong, sorry

4:29 AM
@LuckyChouhan Not yet
@LuckyChouhan Full of mathematics

@TedShifrin nvm, I got it. Thanks.

5:06 AM
What are the fundamental axioms of logic, given the inference rule of modus pones is taken as well used to interpret most of modern mathematics

I highly recommend you check out book of proof. It's what we're using in my intro to math reasoning class @FraserJames
it's free
based off the question you posed on the main site, I think it'd be a great introduction to get you started for analysis, abstract algebra, etc

5:28 AM
@oscarmetalbreak I see $5$, not $7$, conjugacy classes in the second case. Help!

5:52 AM
you are very generous with your time!

I thought I saw it easily, but not. I have a similar exercise in my algebra book.
Are you and your son finished smashing cars?

don't start me! still in the shop. apparently waiting for some part being shipped from another country. over 3 months since he was rear ended!
it is a bit of an issue. i will need a car soon, i'm getting my hip replaced in december, so using my bike for transport won't be an option for a while.

Still covid supply chain messes …

one supposes, if one was generous :-)

December is imminent.

5:58 AM
first world issues, but a pain nontheless

the guy from the shop is driving it right now. doing donuts on a section of 580 they took over just for the purpose of abusing your car.

yep, frankly i'm scared.
:-)

Well, lots to be scared of these days :(

unfortunately i got rid of my wrx, which would be the fun car to drive
yep, recent actions in my birth country have left me a bit depressed

Yeah, and in Burlington, Vt and ….

6:00 AM
i think i am just too naïve

Racism, antisemitism, anti-islam everywhere … Democracy teetering. I’m such an optimist …

sickening. for a while i had hopes for Ireland. they managed to havea gay non-white for pm, i took that as a good sign.

Munchkin will save us.

:-). humans are a sad lot.
and the supply of (non geometry) convex problems has dwindled...
i tried to get my daughter to get her own insurance. but all the agencies she contacted are not taking new clients until the new year. except one, and that was close to $2k for a 6 month period. on the drive to school this morning munchkin said "you said someone died? henry?" because she overheard me telling my wife about kissinger the previous day and i guess had been stewing about the unknown ever since. i was like, oh wow, i definitely can't explain this. 6:14 AM especially the part about theranos i said he was some guy who lived to 100, which then changed the subject of how you live to 100. i reminded her of her mom's uncle, whom she'd met on thanksgiving, and pointed out that he was 90. so if you want to know who henry kissinger is, just take uncle ralph and add 10 years. ;-). my sister sent me a book on ageing (sort of a joke) (i think). it is all about blue zones and living the happy life, so it has no applicability to me. but the day before i read it, i read an article that said a lot of the blue zone longevity came from poor memory & record keeping. huh? is blue zone a new strain of weed? :-) some where in japan, and those cheese eating Euro countries apparently there is even an eponymous show... yeah, you just need to outlive anyone who worked at the hospital, and then the hall of records. or be born before there were any records, like ted. 6:21 AM Life in caves was easier. speaking of records, i still have my tubes lp white punks on dope 6:46 AM if we look at cylinder with ends closed, then we can say that the second fundamental form at point p on it is either 0 or -ve. right? This is because if we orient it norm emerging out of the cylinder, then at each point p the surface bends away from the normal. (or doesn't bend at all, like, at the ends) ends closed how? like you put hemisphere caps on it? not like some kind of orbifold thing? but changing the orientation inward, the fundamental form will be either 0 or positive. @leslietownes I mean like put two coins or lids on both ends of a cylinder. so you're OK with putting a weird cornery edge in it? [using the technical terms] You’re talking about a non-smooth surface. ohh, no edge. ohh so that won't be smooth anymore. 6:52 AM The orientation changes the sign of the second fundamental form. The normal curvatures of a cylinder are$0$and$\pm 1/r$. That ain’t no cylinder. I was thinking of the end at the right end. There is no end. but then it is also not smooth, right? Because it also has a corner. 6:54 AM A whole curve of corners. a weird cornery edge, as it were Why are you going to this? ok so I suppose the image is extending to infinity at the right end in which case we get that the second fundamental form at a point q is either positive or negative. That is, indefinite. Yes, that surface is all saddle points. I have one confusion. 6:57 AM like that texan phrase, "all saddle and no horse" You could send for a monkey. The description under the image says: the second fundamental form at p (i.e. the Weingarten map/shape operation at p) is indefinite. indefinite means neither positive definite nor negative definite. Yes. It curves both inward and outward at each point. You should really read my book. But I understand from the image that since at p the bend is towards N(p) so the fundamental form at p is >0 hence positive definite. @TedShifrin ok You have to look at all normal slices. Not just the curve that gives the profile . 7:01 AM otoh what I say here is for varying p: I was thinking that at the left end, the bend is away from the normal there so the fundamental form there is <0 and in the middle it is positive. 5 mins ago, by Koro ok so I suppose the image is extending to infinity at the right end in which case we get that the second fundamental form at a point q is either positive or negative. Look at the circles. looking at circles, I should get positive definite no? No. They’re bending away from the normal. curves going into and out of the page are bending in a different way than the cross section curve drawn from left to right on the page ohh right, they are bending from the normal. 7:03 AM Look at section 2 of chapter 2 of my text. Lots of pictures, too. the circles are frowning :( the page curve is smiling I see. So looking at the page gives me positive definite but looking at the circles gives negative. well, 'definite' or not would apply to the form, no? you're just talking about the signs of particular curvatures. yes to the form at p. I know that the form is positive definite iff all curvatures are positive. 7:25 AM @TedShifrin thanks a lot. Fig. 2.2 gives details. @leslietownes yes, it makes sense. Thanks a lot. 1 hour later… 8:30 AM I have one confusion now: I take a cylinder defined by level curve of$f(x_1, x_2, x_3)= x_1^2+x_2^2$. Let's take a point on its equator, say p. I have calculated that the second fundamental form$II_p(v)= -v_1^2-v_2^$, where$v= (v_1,v_2,v_3)$. That is, the form is negative definite. But if we look at the vertical line through p, then clearly along this direction the cylinder doesn't bend with respect to the normal at p. So as per this observation, shouldn't the form be 0 here? Thus concluding that the form is semi-definite. and not definite. i.e., from the diagram it is observed that the form is negative along e_1 but 0 along e_2. But this doesn't tally with the formula I got above. 8:54 AM there is no problem.$II_p(0,0,v_3)=0$so all we can say is that the form is semi definite. the form is not definite. 9:54 AM Hi @LuckyChouhan I have cleared the written test today, i.e. I am eligible for the interview 10:30 AM If$f$is a continuous function such that$|f(x)-f(y)|>=ln(1+|x-y|)$then which of the following options are true. A) f is injective. B) f is surjective C) f is strictly increasing or decreasing. A is true, easy to check C follows from objectivity + continuous Injectivity* 10:49 AM @SoumikMukherjee I see @SoumikMukherjee Congratulations! I am happy for you. Thanks @LuckyChouhan @SoumikMukherjee :) have you ever noticed that Ted, Copper-Hat and Leslie they dicussed totally different stuff? @SoumikMukherjee do you remember any problem which you found very interesting in today's test? @LuckyChouhan What do you mean by different stuff? Advanced maths? @leslietownes I meant stuff like this haha @TedShifrin and this 1 hour later… 12:23 PM -1 Let $$f(x) = \sum_{n=1}^{\infty} (-1)^{n+1} a_n x^n$$ where$a_n > 0$and For all real$x>0$the only zero's of$f(x)$are at the positive integers. How fast can$|f(x)|$grow for$x>0$? someone downvoted anything wrong ? and a vote for close ? Is it a dumb question ? 12:55 PM @mick you have quite good reputations score and you're asking such question and not following MSE guidelines you didn't even show your effort. 1:12 PM I found an answer. It is trivial. Should i close or delete ? It was a brain fart i guess just delete it :) Ok @SoumikMukherjee from R to R? Hello Jack 1:33 PM hi all 2:19 PM @Jakobian yes Let$f$be differentiable and bounded at$[1,\infty)$. Suppose that$f'$is continues at$[1,\infty)$. Prove the improper integral$\int_1^\infty f'(x^2)dx$converges I am not sure what I am missing but I am kinda not going anywhere, can someone hint me? Let$I(\lambda)=\int_1^2e^{x^2}e^{-i\lambda x^2}dx$then is$I(\lambda)$differentiable on R? And what is the value of$I(\lambda)$when$\lambda$tends to infinity? I am on my mobile so there may be some latex issues @TedShifrin I thought 5 at the beginning, but then I found out$(x-p)(x+p)$are the two other minimal polynomial and in the case of$GL_2(F)$, they are unique representatives to the conjugacy classes. 3:10 PM @X4J$\int_1^\infty f'(x^2) dx= \int_1^\infty \frac{f'(y)}{\sqrt y} dy=\sum_{n=1}^\infty \int_n^{n+1} \frac{f'(y)}{\sqrt y}$. @mick The reasons that someone might have downvoted have been explained to you innumerable times, both in chat and in response to your questions. The questions that you ask tend to either be completely contextless (which is the case with the one you linked), or long, discursive, ramble-y messes. I would very much encourage you to actually listen to and implement the advice that many people have given you in the past. As I have said before, you need to sell your questions. Explain to other people why they are interesting and why they should care about the answer. For a lot of people, the motivation is something like "I am taking an introduction to calculus class, and was assigned this problem. It is from [text], in [section x], where we have just learned [tool]. I suspect that [tool] should be used, but it is not clear to me how." Hello @XanderHenderson do you know Yitang Zhang? @XanderHenderson ! Richard hamming also said this in his lecture. Most of your questions seem to have something to do with your own particular interests, which means that you actually need to do a better job of selling the work. We can all understand why a student might be motivated to finish a homework problem. I am not particularly compelled by such motivation, but I understand it, at least. Why should anyone care about your questions? @LuckyChouhan No? Should I? @XanderHenderson Really, I am so sad. He proved that largest gap between 2 primes can be < 7*10. @XanderHenderson Yeah why not. Now you have to apply for absolution. @LuckyChouhan I have very little interest in number theory, and have very little knowledge of the people who work there. @LuckyChouhan Yeah, I'm ignoring you, now. 3:24 PM @XanderHenderson hehe, okay have a good day. chat.stackexchange.com/transcript/message/64794161#64794161 oh I meant$7 * (10)^{7}$. I just realized. @Koro Thanks mate @SoumikMukherjee yes and$\int_1^2 e^{x^2}\mathrm{d}x$not$\int_1^2 e^{x^2}\mathrm{d}x$, sorry Let$u = -x^2$then$I(\lambda) = \int_{-4}^{-1} \frac{1}{2\sqrt{-u}}e^{-u} (\cos(\lambda u)+i\sin(\lambda u))\mathrm{d}u$and now using Riemann-Lebesgue lemma this converges to$0$3:52 PM @oscarmetalbreak But the characteristic polynomial must divide$x^4-1$, which has distinct roots. No? @Koro You need to work in local coordinates on the surface, not in$\Bbb R^3$coordinates. @TedShifrin Yes and when$p=4k+1$, one can write$x^4-1=(x^2-1)(x^2+1)=(x-p)(x+p)(x-1)(x+1)$where$p, -p$are the two roots I found for$x^2+1$in this case. @Jakobian Thank you 4:08 PM Don’t use$p$here. It’s already the prime. I still don’t understand how the minimal polynomial can be$x-a$. Then the characteristic polynomial would have to be$(x-a)^2$. Hey @Ted Oh, wait. The minimal polynomial must divide$x^4-1$, not the characteristic polynomial. Hi Lukas! Is there an "intrinsic" definition of real analytic spaces and analytification of finite type schemes over R? The definition I found is: take complexification, then complex analytification, then take Galois invariants feels unsatisfying somehow I mean a matrix in this case could be$\begin{pmatrix} a & 0 \\ 0& a \end{pmatrix}$As you know, I know nothing about this, Lukas. Right, @oscar. I messed up, as I said above. Well done. 4:19 PM why can't you do it "per hand" similar to the complex case Let$f(x)=\sum_{n=1}^{\infty} \frac{nx-[nx]}{n^2}$is$f$continuous at the rationals or the irrationals and is it integrable on a bounded closed subset of R? I have done the continuity part but stuck at the integrable part @Thorgott that's a good question. I'm mostly confused by the fact that a paper uses the construction I described above Riemann integrable I would suppose it's just a lot more convenient to do that then to re-do the entire analytification process over$\mathbb{R}$instead of$\mathbb{C}$it should probably also imply some formal properties of this construction more directly inherited from the complex one hmm, you're probably right 4:33 PM It's my birthday today My age is now a factorial Not saying which one Well,$3! = 6$, and you are likely older than that, while$5!=120$, and I suspect that you are not quite that old. So... Then 2? Well 2 is not older than 6 though Funny that there were 10 questions from analysis and 10 from linear algebra and I remember most of the analysis questions but barely any linear algebra questions @AkivaWeinberger Happy birthday, old man! How is it being 6? @AkivaWeinberger Happy Birthday All I remember that the linear algebra part was flooded with symmetric positive definite matrix questions happy birthday, Akiva 4:42 PM @ Happy Birthday Alles Gute zum Geburtstag @Akiva My bones make noises when I move I always wonder one question, do people feel bad of being old? And how old they need to be so they will feel bad? Careful. I’m the antique one in the crowd. @oscarmetalbreak Old enough to learn that math is not just counting 4:51 PM @TedShifrin Sorry. Didn't mean to. @SoumikMukherjee true @AkivaWeinberger perhaps you're living an inactive lifestyle which made your bones weak @ Jakobian Any hint about the question above? I have a question. For a finite field$F$of characteristic$p$a prime. Will someone end up to be the same field after finitely many times of taking frobineus endomorphism to$F$. For example,let$\phi$be the endomorphism, for every$n>N$, then$\phi^n(F)$will always be the same. If so what could it be? I am guessing if it could be the prime ring. @SoumikMukherjee what question you're asking too many questions and I'm not sure which one @oscarmetalbreak$\phi$is bijective for every finite field 4:59 PM a finite field$F$of characteristic$p$has$p^n$elements for some$n$and then the$n$-th power of the Frobenius will be trivial ah, you're only asking about the image, refer to Lukas' comment 38 mins ago, by Soumik Mukherjee Let$f(x)=\sum_{n=1}^{\infty} \frac{nx-[nx]}{n^2}$is$f$continuous at the rationals or the irrationals and is it integrable on a bounded closed subset of R? @Jakobian about the integrable part @LukasHeger Isn't that the frobenius endomorphism could be injective? And so the image would be a subring? @SoumikMukherjee okay so$nx-[nx] = \{nx\}$, and$|\{nx\}|\leq 1$. Use Weierstrass M-test well, it's an injective map from a finite field to itself, so it's bijective @SoumikMukherjee Since$f$is measurable and bounded, its integrable on any bounded subset 5:05 PM I didn't have the question formulated well since it is just a vogue question I thought yesterday when I looked at the wiki of frobenius endomorphism. by$\phi^n(F)$, you're asking about the image of$\phi^n$? ah right I suppose it would be more interesting to ask whether the descending chain of subfields obtained by repeated application of the Frobenius ever stabilizes for non-perfect fields @Jakobian Oh okay, thanks again Yes I am asking the image of$\phi^n(F)\$.

5:09 PM
(the answer to my refined question is "no")

@Thorgott Will this be covered in the study of Galois theory?

@Thorgott yeah. The perfection has always infinite degree, because if a finite extension of a field is finite, it is itself perfect

the Frobenius endomorphism? perfect fields? most likely, yes
the specific question of whether the chain of subfields by repeated applications of the Frobenius stabilizies? no, but you can answer that question as an exercise
@LukasHeger calling it the "perfection" is great

I was too lazy to write "colimit over powers of frobenius"

5:31 PM
@AkivaWeinberger Happy 4! birthday :)

5:45 PM
Hi :) I just tried cutting my own hair. It's a disaster! I look like Gollum! That is all . . .

@Shaun I cut my own hair every five or six days.

I need to get into that habit, @XanderHenderson. I'm balding :/
I think it's funny, it's alright.
This has nothing to do with maths, of course, but I thought it'd make some of you laugh.
This is my kind of social media.

@Shaun I have no idea whether I am balding or not. I've been shaving my head since my early 20s. I went from hair down to my ass to shaved bald in an evening.
(I started shaving my head when I was fencing competitively---my hair kept getting caught in the velcro on the mask, and it seemed that no hair would solve that problem and be much easier to maintain).

Vote for Pedro!
@XanderHenderson That's cool, and very pragmatic!

6:04 PM
@Shaun would you please send us the photo?
@XanderHenderson how is this possible?

@LuckyChouhan Us?

@SoumikMukherjee Soumik you remain online all the time
@SoumikMukherjee yeah :))
don't you want to see his new haircut ?

19 mins ago, by Xander Henderson
@Shaun I have no idea whether I am balding or not. I've been shaving my head since my early 20s. I went from hair down to my ass to shaved bald in an evening.

Because you cut them before they grow.

That is the implication, yes.

6:14 PM
@XanderHenderson now I see your profile photo is really you :(

@LuckyChouhan No chance!

Do you cut your eyebrows too?

Looks like the House expelled George Santos.

@Shaun please we want to see :)

@LuckyChouhan No! I've been on the internet long enough to know that's a mistake! :p

6:16 PM
@LuckyChouhan we?

@Shaun you once did 1000 days on MSE. So how did you feel? Did you think that oh almost 3yrs has passed and what was my progress something like that
@SoumikMukherjee sucks

@LuckyChouhan It just sort of happened. I noticed the counter was close one day, so decided to make a note of it. It's upwards of 1,500 these days.

@Shaun cool, do you use programming in your math?

I do.
It's in GAP exclusively.
I'm in the process of starting to use a supercomputer/HPC.

great, whenever I need to compute something time consuming I use WolframAlpha :)

6:22 PM
@XanderHenderson Awww.
2

@TedShifrin I know, right? Couldn't have happened to a nicer guy...
2

I have no idea if he’s ever told the truth. Is he actually gay? Possibly.

@XanderHenderson do you want Trump to be president again?

@TedShifrin Who knows. I don't even care. Just happy he's gone.

6:24 PM
@LuckyChouhan WolframAlpha doesn't have the capabilities of GAP, though, I'm afraid.

My point, being gay myself, was that he used this as one of many ways to claim to be down-trodden.

@Shaun yeah, actually I use whenever I do some number theory stuff. I just saw GAP is amazing.
@TedShifrin what?? Are you gay?

I just said so. Pay attention.
5

You don't have to overreact after anything you find shocking

6:28 PM
@SoumikMukherjee Please tell me also haha

its pretty rude what you did
2

I made this clear in this chatroom many years ago. Now we move on …

oh sorry Ted,

Unlike Santos, I don’t lie. And I actually attended and graduated from a university without being a fake volleyball star.
3
I also don’t commit federal crimes by lying repeatedly to the government or billions of citizens.
3

@TedShifrin but why someone asked you that? I never asked someone question about their sexuality even I do lots of nonsense.

6:33 PM

No one asked. I volunteered the info. It’s important to me to be a different sort of role model … as I always tried yo be for my students.
Yes, and his mom who was killed on 9/11.

@TedShifrin Well, millions. There are not a billion US citizens, so far as I know.

Is Santos this politician from Florida I heard about

@Jakobian No, he's in New York, I think.
Desantis is in Florida.

Ah, yeah. I mixed them up
Wikipedia is fast, they already have that he's expelled
Interesting. So he's a liar and a thief (according to wikipedia)

6:40 PM
He's been at it for almost exactly a year now?
That's quick clean-up.
@XanderHenderson NY 3rd, yes.

@Jakobian "Allegedly".

Not sure how relevant 'allegedly' is when they've publicly admitted the lying part.
It's easier to find dirt on Santos than on Trump. No joke.
That's saying something.
> Among other things, Santos told the New York Post that he had not worked “directly” for Goldman Sachs and Citigroup, saying that a company he did work for did business with both of them.

He also said he had not graduated from Baruch College, nor “from any institution of higher learning.”
They tried to expel him before, so I'm surprised they finally managed anyway.
Not that I'm trying to crash anyone's party, circumstances brought me here.

Circumstances?

Mysteeeeeerious circumstances!
*wiggles fingers*

@Xander I guess I just proved that I do lie — unwittingly.

6:50 PM
@TedShifrin Oh noes!

Pops fresh popcorn for everybody :-D

I'm here to hear Xander talk about rectangles.

Rectangles are pointy. But not too pointy.

🍿🍿🍿🍿🍿🍿

@XanderHenderson Well said.

6:57 PM
expelling desantis would be too good to be true

There's too many thugs in politics these days.

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